Calculus 3: Work Homework: Electric Dipole Field

In summary, An electric dipole with dipole moment p = 4 × 10−5 C-m sets up an electric field given by F(x,y,z) = kp/r^5 <3xz, 3yz, 2z^{2} - x^{2} - y^{2}>, with r = (x^{2} + y^{2} + z^{2})^{1/2} and k = 8.99 × 109 N-m^{2}/C^{2}. The work required to move a particle of charge q = 0.01 C from (1,−5, 0) to (3, 4, 4) against the force is given by the
  • #1
tarantino5
1
0

Homework Statement



An electric dipole with dipole moment p = 4 × 10−5 C-m sets up
an electric field (in Newtons per coulomb)

F(x,y,z) = [itex]kp/r^5[/itex] <3xz, 3yz, 2z[itex]^{2}[/itex] - x[itex]^{2}[/itex] - y[itex]^{2}[/itex]>

where r = (x[itex]^{2}[/itex] + y[itex]^{2}[/itex] + z[itex]^{2}[/itex])[itex]^{1/2}[/itex] with distance in meters and k = 8.99 ×
109 N-m[itex]^{2}[/itex]/C[itex]^{2}[/itex]. Calculate the work against F required to move a particle
of charge q = 0.01 C from (1,−5, 0) to (3, 4, 4). Note: The force on
q is qF Newtons.

Homework Equations



∫[itex]^{b}_{a}[/itex] F(c(t)) * c'(t) dt

The Attempt at a Solution



I know that I need to set ∇f = F and find a potential function. I'm not sure what the parametric equation should be.
 
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  • #2
tarantino5 said:
I know that I need to set ∇f = F and find a potential function. I'm not sure what the parametric equation should be.

Perhaps you don't need to calculate the potential at all. Seems like you get work directly from the force, according to your formula.

The path parametrization depends on the path of course. It seems a path is not specified in the problem. Why is it not needed to solve the problem?
 

FAQ: Calculus 3: Work Homework: Electric Dipole Field

What is Calculus 3 and how does it relate to electric dipole field?

Calculus 3, also known as multivariable calculus, is a branch of mathematics that deals with the study of functions of multiple variables. It is often used in physics to analyze the behavior of systems with multiple variables, such as electric fields. In the case of electric dipole field, Calculus 3 is used to calculate the work done by the field on a charge.

How is work related to the electric dipole field?

Work is related to the electric dipole field through the concept of potential energy. The electric dipole field exerts a force on charges, causing them to move from one point to another. The work done by the field is equal to the change in potential energy of the charge. This is calculated using the formula W = -ΔU, where W is work, ΔU is the change in potential energy, and the negative sign indicates that the work is done by the field on the charge.

What is the formula for calculating the electric dipole field?

The formula for calculating the electric dipole field is E = k(Q/r^2)cosθ, where E is the electric field strength, k is a constant, Q is the charge of the dipole, r is the distance from the dipole, and θ is the angle between the dipole axis and the field point. This formula takes into account both the magnitude and direction of the field at a given point.

How is the electric dipole field different from other types of electric fields?

The electric dipole field is different from other types of electric fields because it is created by two equal and opposite charges that are separated by a small distance. This creates a distinct pattern of field lines that can be visualized as two arrows pointing in opposite directions. In contrast, other types of electric fields may have various configurations of charges and therefore, a more complex pattern of field lines.

How is Calculus 3 used to solve problems involving electric dipole field?

Calculus 3 is used in solving problems involving electric dipole field by providing the necessary mathematical tools to calculate the work done by the field on a charge. This involves integrating the electric field strength over a given path to find the potential energy, and then taking the negative of that value to find the work done. Additionally, Calculus 3 can also be used to find the direction and magnitude of the electric field at a given point by taking partial derivatives of the field formula.

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